Problem: Divide the following complex numbers. $ \dfrac{-6+2i}{1+3i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1-3i}$ $ \dfrac{-6+2i}{1+3i} = \dfrac{-6+2i}{1+3i} \cdot \dfrac{{1-3i}}{{1-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-6+2i) \cdot (1-3i)} {(1+3i) \cdot (1-3i)} = \dfrac{(-6+2i) \cdot (1-3i)} {1^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-6+2i) \cdot (1-3i)} {(1)^2 - (3i)^2} = $ $ \dfrac{(-6+2i) \cdot (1-3i)} {1 + 9} = $ $ \dfrac{(-6+2i) \cdot (1-3i)} {10} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-6+2i}) \cdot ({1-3i})} {10} = $ $ \dfrac{{-6} \cdot {1} + {2} \cdot {1 i} + {-6} \cdot {-3 i} + {2} \cdot {-3 i^2}} {10} $ Evaluate each product of two numbers. $ \dfrac{-6 + 2i + 18i - 6 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{-6 + 2i + 18i + 6} {10} = \dfrac{0 + 20i} {10} = 2i $